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Maths 4 Marks

Probability · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 275 questions.

Questions · Page 4 of 6

4 Marks

Question 1514 Marks
Coloured balls are distributed in four boxes as shown in the following table:
Box
Colour
Black
White
Red
Blue
I
II
III
IV
3
2
1
4
4
2
2
3
5
2
3
1
6
2
1
5
A box is selected at random and then a ball is randomly drawn from the selected box. The colour of the ball is black, what is the probability that ball drawn is from the box III.
Answer
Let A, E1, E2, E3 and E4 denote the events that the ball is black, box I selected, box II selected, box III is selected and box IV is selected respectively.
$\therefore\ \text{P}(\text{E}_1)=\frac{1}{4}$
$\text{P}(\text{E}_2)=\frac{1}{4}$
$\text{P}(\text{E}_3)=\frac{1}{4}$
$\text{P}(\text{E}_3)=\frac{1}{4}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=\frac{3}{18}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{2}{8}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=\frac{1}{7}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_4}\Big)=\frac{4}{13}$
Using Bayes' theorem, we get
Required probability $\text{P}\Big(\frac{\text{E}_3}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{1}{4}\times\frac{1}{7}}{\frac{1}{4}\times\frac{3}{18}+\frac{1}{4}\times\frac{2}{8}+\frac{1}{4}\times\frac{1}{7}+\frac{1}{4}\times\frac{4}{13}}$
$=\frac{\frac{1}{7}}{\frac{1}{6}+\frac{1}{4}+\frac{1}{7}+\frac{1}{13}}=\frac{156}{947}$
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Question 1524 Marks
One bag contains 4 yellow and 5 red balls. Another bag contains 6 yellow and 3 red balls. A ball is transferred from the first bag to the second bag and then a ball is drawn from the second bag. Find the probability that ball drawn is yellow.
Answer
Bag I contains 4 yellow and 5 red balls
Bag II contains 6 yellow and 3 red balls
Transfer can be done in two ways:
I - A red ball is transferred from bag I to bag II and then one yellow ball is drawn from bag II.
II - A red ball is transferred from bag I to bag II and then one yelllow ball is drawn from bag II.
Let E1, E2 and A be events as:
E1 = One yellow ball drawn from bag I
E2 = One red ball drawn from bag I
A = one yellow ball draw from bag II.
$\text{P}(\text{E}_1)=\frac{4}{9}$
$\text{P}(\text{E}_2)=\frac{4}{9}$
$\text{P}(\text{A}|\text{E}_1)=\frac{7}{10}$
[Since E1 has increased one yellow ball in bag II]
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{6}{10}$
[Since E2 has increased one red ball in bag II]
By law of total probability,
$\text{P(A)}=\text{P}(\text{E}_1)\text{ P}(\text{A}|\text{E}_1)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)$
$=\frac{4}{9}\times\frac{7}{10}\times\frac{5}{9}\times\frac{6}{10}$
$=\frac{28+30}{90}$
$=\frac{58}{90}$
$=\frac{29}{45}$
Required probability $=\frac{29}{45}$
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Question 1534 Marks
The probability that A hits a target is $\frac{1}{3}$ and the probability that B hits it, is $\frac{2}{5}$, What is the probability that the target will be hit, if each one of A and B shoots at the target?
Answer
Given,
Probability that A hits a target $=\frac{1}{3}$
$\Rightarrow\ \text{P(A)}=\frac{1}{3}$
Probability that B hits the targer $=\frac{2}{5}$
$\Rightarrow\ \text{P(B)}=\frac{2}{5}$
P (Target will be hit)
= 1 - P (target will not be hit)
= 1 - P (Niether A non B hits the target)
$=1-\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$
$=1-\text{P}(\overline{\text{A}})\text{P}(\overline{\text{B}})$
$=1-[1-\text{P(A)}][1-\text{P}(\overline{\text{B}})]$
$=1-\Big[1-\frac{1}{3}\Big]\Big[1-\frac{2}{5}\Big]$
$=1-\frac{2}{3},\frac{3}{5}$
$=1-\frac{2}{5}$
$=\frac{2}{5}$
Required probability $=\frac{2}{5}$
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Question 1544 Marks
An urn contains 5 red and 2 blcak balls. Two balls are randomly drawn, without replacement. Let X represent the number of black balls drawn. What are the possible values of X? Is X a random variable? If yes, then find the mean and variance of X.
Answer
X can assume values 0 to 2.

Yes X is a random variable.

P(X = 0) = (Probability of getting no black ball)

$=\frac{\text{}^{2}\text{C}_0\times\text{}^{5}\text{C}_2}{\text{}^{7}\text{C}_2}=\frac{1\times\frac{5\times4}{2\times1}}{\frac{7\times6}{2\times1}}=\frac{20}{42}$

P(X = 1) = (Probability of getting one black ball)

$=\frac{\text{}^{2}\text{C}_1\times\text{}^{5}\text{C}_1}{\text{}^{7}\text{C}_2}=\frac{1\times5}{\frac{7\times6}{2\times1}}=\frac{20}{42}$

P(X = 2) = (Probability of getting two black balls)

$=\frac{\text{}^{2}\text{C}_2\times\text{}^{5}\text{C}_0}{\text{}^{7}\text{C}_2}=\frac{1\times1}{\frac{7\times6}{2\times1}}=\frac{2}{42}$

Thus, Probability distribution of random variable X is

$\text{X}$
$0$
$1$
$2$
$\text{P}(\text{X}) $
$\frac{20}{42}$
$\frac{20}{42}$ 
$\frac{2}{42}$

 

$\text{x}_\text{i}$

 $\text{p}_\text{i}$

 $\text{p}_\text{i}\text{x}_\text{i}$

 $\text{p}_\text{i}\text{X}_\text{i}^2$
$0$

 $\frac{20}{42}$

 $0$

 $0$
$1$

 $\frac{20}{42}$

 $\frac{20}{42}$

 $\frac{20}{42}$
$2$

 $\frac{2}{42}$

 $\frac{4}{42}$

 $\frac{8}{42}$

 

 

 $\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{4}{7}$

 $\sum\text{p}_\text{i}\text{x}_\text{i}^2=\frac{2}{3}$

Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{4}{7}$

Variance $=\sum\text{p}_\text{i}\text{x}_\text{i}^2-\big(\sum\text{p}_\text{i}\text{x}_\text{i}\big)^2$

$=\frac{2}{3}-\Big(\frac{4}{7}\Big)^2=\frac{50}{147}$

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Question 1554 Marks
A bag contains 3 white and 2 black balls and another bag contains 2 white and 4 black balls. One bag is chosen at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is white.
Answer
Bag I contains 3 white and 2 black balls
Bag II contains 2 white and 4 black balls
One bag is chosen at random, then one ball is drawn and its is while.
Let E1, E2 and A be events as:
E1 = Selecting bag I
E2 = Selecting bag II
A = Drawing one white ball
$\text{P}(\text{E}_1)=\frac{1}{2}$
$\text{P}(\text{E}_2)=\frac{1}{2}$
[Since there are only 2 bags]
P(A|E1) = P(Drawing a white ball from bag I)
$=\frac{3}{5}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P}$ [Drawing a white ball from bag II]
By law of total probability,
$\text{P(A)}=\text{P}(\text{E}_1)\text{ P}(\text{A}|\text{E}_1)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)$
$=\frac{1}{2}\times\frac{3}{5}+\frac{1}{2}\times\frac{2}{6}$
$=\frac{3}{10}+\frac{2}{12}$
$=\frac{18+10}{60}$
$=\frac{28}{60}$
$=\frac{7}{15}$
Required probability $=\frac{7}{15}$
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Question 1564 Marks
A and B are two independent events. The probability that A and B occur is $\frac{1}{6}$ and the probability that neither of them occurs is $\frac{1}{3}$. Find the probability of occurrence of two events.
Answer
Given
$\text{P}(\text{A}\cap\text{B})=\frac{1}{6}$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\frac{1}{3}$
We know that,
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\text{P(A) }\text{P(B)}$
$\frac{1}{3}=(1-\text{P(A)})(1-\text{P(A)})$
$\frac{1}{3}=1-\text{P(B)}-\text{P(A)}+\text{P(A) }\text{P(B)}$
$\frac{1}{3}=1-\text{P(B)}-\text{P(A)}+\text{P}(\text{A}\cap\text{B})$
$\frac{1}{3}=1-\text{P(B)}-\text{P(B)}+\frac{1}{6}$
$\text{P(A)}+\text{P(A)}=\frac{1}{1}+\frac{1}{6}-\frac{1}{3}$
$=\frac{6+1-2}{6}$
$\text{P(A)}+\text{P(B)}=\frac{5}{6}$
$\text{P(A)}=\frac{5}{6}-\text{P(B)}\ .....\text{(i)}$
Given, $\text{P}(\text{A}\cap\text{B})=\frac{1}{6}$
$\text{P(A) } \text{P(B)}=\frac{1}{6}$
$\Big[\frac{5}{6}-\text{P(B)}\Big]\text{P(B)}=\frac{1}{6}$
[Using equation (i)]
$\Rightarrow\ \frac{5}{6}\text{P(B)}-\big\{\text{P(B)}\big\}^2=\frac{1}{6}$
$\Rightarrow\ \{\text{P(B)}\}^2-\frac{5}{6}\text{P(B)}+\frac{1}{6}=0$
$\Rightarrow\ 6\{\text{P(B)}\}^2-5\text{P(B)}+1=0$
$\Rightarrow\ 6\{\text{P(B)}\}^2-3\text{P(B)}-2\text{P(B)}+1=0$
$\Rightarrow\ 3\text{P(B)}[2\text{P(B)}-1]-1[2\text{P(B)}-1]=0$
$\Rightarrow\ [2\text{P(B)}-1][3\text{P(B)-1}]=0$
$\Rightarrow\ 2\text{P(B)}-1 = 0\text{ or }3\text{P(B)}-1=0$
$\Rightarrow\ \text{P(B)}=\frac{1}{2} \text{ or P(B)}=\frac{1}{3}$
⇒ Using equation (i),
$\text{P(B)}=\frac{1}{2}\Rightarrow\ \text{P(A)}=\frac{5}{6}-\frac{1}{2}=\frac{1}{3}$
$\text{P(B)}=\frac{1}{2}\Rightarrow\ \text{P(A)}=\frac{5}{6}-\frac{1}{3}=\frac{1}{2}$
Hence, $\text{P(B)}=\frac{1}{2},\text{P(A)}=\frac{1}{3} \text{ or }\text{P(B)}=\frac{1}{3},\text{P(A)}=\frac{1}{2}$
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Question 1574 Marks
A bag contains 3 red and 2 black balls. One ball is drawn from it at random. Its colour is noted and then it is put back in the bag. A second draw is made and the same procedure is repeated. Find the probability of drawing,
  1. Two red balls,
  2. Two black balls,
  3. First red and second black ball.
Answer
Given bag contains 3 red and 2 black balls.

A = Getting one red ball

$\Rightarrow\ \text{P(A)}=\frac{3}{5}$

B = Getting one black ball

$\Rightarrow\ \text{P(B)}=\frac{2}{5}$

  1. P(Getting two red balls)

= P(A) P(A)

$=\frac{3}{5}\times\frac{3}{5}$

$=\frac{9}{25}$

P(Getting two red balls) $=\frac{9}{25}$

  1. P(Getting two black balls)

= P(B) P(B)

$=\frac{2}{5}\times\frac{2}{5}$

$=\frac{4}{25}$

P(Getting two black balls) $=\frac{4}{25}$

  1. P(Getting first red and second black ball)

= P(A) P(B)

$=\frac{3}{5}\times\frac{2}{5}$

$=\frac{6}{25}$

P(Getting first red and second black ball) $=\frac{6}{25}$

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Question 1584 Marks
cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the mean and variance of red cards.
Answer
It is given that the cards are drawn successively with replacement so the events are indepent.

Therefore, the drawing of the cards follow binomial distribution.

Probability of drawng a red card $=\text{p}=\frac{26}{52}=\frac{1}{2}$

$\therefore\text{q}=1-\text{p}=1-\frac{1}{2}=-\frac{1}{2}$

Also, $\text{n}=3$

Let X be the random variable denoting the number of red cards drawn from a well shuffied pack of 52 cards.

$\therefore\text{P(X = r})=\text{ }^3\text{C}_{\text{r}}\big(\frac{1}{2}\big)^{3-\text{r}}\big(\frac{1}{2}\big)^{\text{r}}=\text{ }^3\text{C}_{\text{r}}\big(\frac{1}{2}\big)^3,\text{r}=0,1,2,3$

Probability of drawing no red ball $=\text{P(X}=0)=\text{ }^3\text{C}_0\big(\frac{1}{2}\big)^3=\frac{1}{8}$

Probability of drawing one red ball $=\text{P(X}=1)=\text{ }^3\text{C}_1\big(\frac{1}{2}\big)^3=\frac{3}{8}$

Probability of drawing tow red balls $=\text{P(X}=2)=\text{ }^3\text{C}_2\big(\frac{1}{2}\big)^3=\frac{3}{8}$

Probability of drawing three red balls $=\text{P(X}=3)=\text{ }^3\text{C}_3\big(\frac{1}{2}\big)^3=\frac{1}{8}$

Thus, the probability distribution of X is as follows:

$\text{x}_{\text{i}}$

 $\text{p}_{\text{i}}$

 $\text{p}_{\text{i}}\text{x}_{\text{i}}$

 $\text{p}_{\text{i}}\text{x}_{\text{i}}^2$
$0$

 $\frac{1}{8}$

 $0$

 $0$
$1$

 $\frac{3}{8}$

 $\frac{3}{8}$

 $\frac{3}{8}$
$2$

 $\frac{3}{8}$

 $\frac{6}{8}$

 $\frac{12}{8}$
$3$

 $\frac{1}{8}$

 $\frac{3}{8}$

 $\frac{9}{8}$

 

 

 $\sum\text{p}_{\text{i}}\text{x}_{\text{i}}=\frac{12}{8}$

 $\sum\text{p}_{\text{i}}\text{x}_{\text{i}}^2=3$
Mean of X $=\sum\text{p}_{\text{i}}\text{x}_{\text{i}}=\frac{12}{8}=\frac{3}{2}$

variance of X $=\sum\text{p}_{\text{i}}\text{x}_{\text{i}}^2-(\text{Mean})^2=3-\frac{9}{4}=\frac{3}{4}$

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Question 1594 Marks
Prove that in throwing a pair of dice, the occurrence of the number 4 on the first die is independent of the occurrence of 5 on the second die.
Answer
A pair of dice are thrown. It has 36 elem ents in its samplw space.
A = Occurence of number 4 on firs die
A = {(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)}
B = Occurence of 5 on second die
B = {(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5)}
$\text{A}\cap\text{B}=\Big\{(4,5)\Big\}$
$\text{P(A)}=\frac{6}{36}=\frac{1}{6}$
$\text{P(B)}=\frac{6}{36}=\frac{1}{6}$
$\text{P}(\text{A}\cap\text{B})=\frac{1}{36}$
$\text{P(A)}.\text{P(B)}\frac{1}{6}\times\frac{1}{6}$
$=\frac{1}{36}$
$\text{P(A)}.\text{P(B)}=\text{P}(\text{A}\cap\text{B})$
So, A and B are indepepndent events.
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Question 1604 Marks
Two natural numbers r, s are drawn one at a time, without replacement from the set S = {1, 2, 3, ......., n}. Find $\text{P}\big[\text{r}\leq\text{p}|\text{s}\leq\text{p}\big],$ where $\text{p }\in\text{ S.}$
Answer
$\because$ set S = {1, 2, 3, ......., n}
$\therefore\text{P}\big(\text{r}\leq\text{p}|\text{s}\leq\text{p})=\frac{\text{P}(\text{p}\cap\text{S})}{\text{P}(\text{S})}$
$=\frac{\text{p}-1}{\text{n}}\times\frac{\text{n}}{\text{n}-1}=\frac{\text{p}-1}{\text{n}-1}$
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Question 1614 Marks
The probability that a student entering a university will graduate is 0.4. Find the probability that out of 3 students of the university.
  1. none will graduate.
  2. only one will graduate.
  3. all will graduate.
Answer
Let X be the number of students that gradute from among 3 students.
Let p = probability that a student entering a university will garduate.
Here, n = 3, p = 0.4 and q = 0.6
Hence, the distribution is given by
$\text{P(X = r})=\text{ }^3\text{C}_{\text{r}}(0.4)^{\text{r}}(0.6)^{3-\text{r}},\text{r}=0,1,2,3$
  1. $\text{P(X}=0)=\text{q}^3=0.216$
  2. $\text{P(X}=1)=3(0.4)(0.36)=0.432$
  3. $\text{P(X}=3)=\text{p}^3=0.064$
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Question 1624 Marks
A die is thrown thrice. Find the probability of getting an odd number at least once.
Answer
P(getting an odd number in one throw) $=\frac{1}{2}$
Here, getting an odd number in three throws refers to 3 independent events.
$\text{P(A)}=\text{P{B}}=\text{P(C)}=\frac{1}{2}$
$\text{P}(\text{A}\cup\text{B}\cup\text{C})=\text{P(A)}+\text{P(B)}+\text{P(C)} \\ -[\text{P}(\text{A}\cap\text{B})+(\text{B}\cap\text{C})+(\text{C}\cap\text{A})]+\text{P}(\text{A}\cap\text{B}\cap\text{C})$
$=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}-\Big[\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}\Big] \\ +\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}$
$=\frac{3}{2}-\frac{3}{4}+\frac{1}{8}$
$=\frac{12-6+1}{8}$
$=\frac{7}{8}$
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Question 1634 Marks
Find the probability distribution of the maximum of the two scores obtained when a die is thrown twice. Determine also the mean of the distribution.
Answer
Let X is the random variable score obtained when a die is thrown twice.
$\therefore$ X = 1, 2, 3, 4, 5, 6
n(S) = 36 [as(1, 1)]
$\therefore\text{P}(\text{X}=1)=\frac{1}{36}$ [as (1, 1)] 
$\text{P}(\text{X}=2)=\frac{3}{36}$ [as (1, 2), (2, 1), (2, 2)]
 $\text{P}(\text{X}=3)=\frac{5}{36}$ [as (1, 3), (2, 3), (3, 1), (3, 2), (3, 3)]
Similarly, $\text{P}(\text{X}=4)=\frac{7}{36},\text{P}(\text{X}=5)=\frac{9}{36},\text{P}(\text{X}=6)=\frac{11}{36}$
So, the required distribution is,
$\text{X}$
 $1$
 $2$
 $3$
 $4$
 $5$
 $6$
$\text{P}(\text{X})$
 $\frac{1}{36}$
 $\frac{3}{36}$
 $\frac{5}{36}$
 $\frac{7}{36}$
 $\frac{9}{36}$
 $\frac{11}{36}$
 $\therefore$ Mean $\left\{\text{E}(\text{X})\right\}=\sum\text{XP}(\text{X})$
$=\frac{1}{36}+\frac{6}{36}+\frac{15}{36}+\frac{28}{36}+\frac{45}{36}+\frac{66}{36}$
$=\frac{161}{36}$
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Question 1644 Marks
From a lot containing 25 items, 5 of which are defective, 4 are choosen at random. Let X be the number of defective found. Obtain the probability distribution of X if the item are chosen without replacement.
Answer
Let X denote the number of defective item in a sample of 4 items drawn from a bag containing 5 defective items and 20 good items. Then, X can take values 0, 1, 2, 3 and 4.
Now,
P(X = 0)
= P(no defective item)
$=\frac{\text{}^{20}\text{C}_4}{\text{}^{25}\text{C}_4}$
$=\frac{4845}{12650}$
$=\frac{969}{2530}$
P(X = 1)
= P(1 defective item)
$=\frac{\text{}^5\text{C}_1\times\text{}^{20}\text{C}_3}{\text{}^{25}\text{C}_4}$
$=\frac{5700}{12650}$
$=\frac{114}{253}$
P(X = 2)
= P(2 defective item)
$=\frac{\text{}^5\text{C}_2\times\text{}^{20}\text{C}_2}{\text{}^{25}\text{C}_4}$
$=\frac{1900}{12650}$
$=\frac{38}{253}$
P(X = 3)
= P(3 defective item)
$=\frac{\text{}^5\text{C}_3\times\text{}^{20}\text{C}_1}{\text{}^{25}\text{C}_4}$
$=\frac{200}{12650}$
$=\frac{4}{253}$
P(X = 4)
= P(4 defective item)
$=\frac{\text{}^5\text{C}_4}{\text{}^{25}\text{C}_4}$
$=\frac{5}{12650}$
$=\frac{1}{2530}$
Thus, the probability distribution of X is given by
$\text{X}$
$0$
$1$
$2$
$3$
$4$
$\text{P}(\text{X})$
$\frac{969}{2530}$
$\frac{114}{253}$
$\frac{38}{253}$
$\frac{4}{253}$
$\frac{1}{2530}$
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Question 1654 Marks
Two groups are competing for the positions of the Board of Directors of a Corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.
Answer
Let E1 and E2 denote the events that the first group and the second group win the competition, respectively. Let A be the event of introducing a new product.
P(E1) = Probability that the first group wins the competition = 0.6
P(E2) = Probability that the second group wins the competition = 0.4
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=$ Probability of introducing a new product if the first group wins = 0.7
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=$ Probability of introducing a new product if the second group wins = 0.3
The probability that the new product is introduced by the second group is given by $\text{P}\Big(\frac{\text{E}_2}{\text{A}}\Big).$
Using Bayes' theorem, we get
Required probability $=\text{P}\Big(\frac{\text{E}_2}{\text{A}}\Big)=\frac{\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{0.4\times0.3}{0.6\times0.7+0.4\times0.3}$
$=\frac{0.12}{0.54}=\frac{2}{9}$
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Question 1664 Marks
If A and B are two events such that,
$\text{P(A)}=\frac{7}{13},\text{P(B)}=\frac{9}{13}$ and $\text{P}(\text{A}\cap\text{B})=\frac{4}{13},$ then find $\text{P}(\overline{\text{A}}|\text{B}).$
Answer
We have,
$\text{P(A)}=\frac{7}{13},\text{P(B)}=\frac{9}{13}$ and $\text{P}(\text{A}\cap\text{B})=\frac{4}{13},$
As, $\text{P}(\overline{\text{A}}\cap\text{B})=\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \text{P}(\overline{\text{A}}\cap\text{B})=\frac{9}{13}-\frac{4}{13}$
$\Rightarrow\ \text{P}(\overline{\text{A}}\cap\text{B})=\frac{9-4}{13}$
$\Rightarrow\ \text{P}(\overline{\text{A}}\cap\text{B})=\frac{5}{13}$
Now,
$\text{P}(\overline{\text{A}}|\text{B})=\frac{\text{P}(\overline{\text{A}}\cap\text{B})}{\text{P(B)}}=\frac{\Big(\frac{5}{13}\Big)}{\Big(\frac{9}{13}\Big)}=\frac{5}{9}$
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Question 1674 Marks
A bag contains (2n + 1) coins. It is known that n of these coins have a head on both sides where as the rest of the coins are fair. A coin is picked up at random from the bag and is tossed. If the probability that the toss results in a head is $\frac{31}{42},$ determine the value of n.
Answer
Given, n coin have head on both sides and (n + 1) coins are fair coins.
Let E1 = Event that an unfair coin is selected.
E2 = Event that a fair coin is selected.
E = Event that the toss results in a head.
$\therefore\text{P}(\text{E}_1)=\frac{\text{n}}{2\text{n}+1}$ and $\text{P}(\text{E}_2)=\frac{\text{n}-1}{2\text{n}+1}$
Also, $\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)=1$ and $\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)=\frac{1}{2}$
$\therefore\text{P}(\text{E})=\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)$
$=\frac{\text{n}}{2\text{n}+1}\cdot1+\frac{\text{n}+1}{2\text{n}+1}\cdot\frac{1}{2}$
$\Rightarrow\frac{31}{42}=\frac{2\text{n}+\text{n}+1}{2(2\text{n}+1)}$
$\Rightarrow\frac{31}{42}=\frac{3\text{n}+1}{4\text{n}+2}$
$\Rightarrow124\text{n}+62=126\text{n}+42$
$\Rightarrow2\text{n}=20\Rightarrow\text{n}=10$
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Question 1684 Marks
Five bad oranges are accidently mixed with 20 good ones. If four oranges are drawn one by one successively with replacement, then find the probability distribution of number of bad oranges drawn. Hence find the mean and variance of the distribution.
Answer
Let X be the random variable denoting the number of bad oranges drawn.
P (getting a good orange) $=\frac{20}{25}=\frac{4}{5}$
P (getting a bad orange) $=\frac{5}{25}=\frac{1}{5}$
The probability distribution of X is given by
$\text{X}$
 $0$
 $1$
 $2$
 $3$
 $4$
$\text{P(X)}$
 $\big(\frac{4}{5}\big)^4=\frac{256}{625}$
 $\text{ }^4\text{C}_1\big(\frac{4}{3}\big)^3\big(\frac{1}{5}\big)=\frac{256}{625}$
 $\text{ }^4\text{C}_2\big(\frac{4}{5}\big)^2\big(\frac{1}{5}\big)^2=\frac{96}{625}$
 $\text{ }^4\text{C}_3\big(\frac{4}{5}\big)\big(\frac{1}{5}\big)^3=\frac{16}{625}$
 $\big(\frac{1}{5}\big)^4=\frac{1}{625}$
Mean of X is given by
$\overline{\text{X}}=\sum\text{P}_{\text{i}}\text{X}_{\text{i}}$
$=0\times\frac{256}{625}+1\times\frac{256}{625}+2\times\frac{96}{625}+3\times\frac{16}{625}+4\times\frac{1}{625}$
$=\frac{1}{625}(256+192+48+4)$
$=\frac{4}{5}$
Variance of X given by
$\text{Var (X)}=\sum\text{P}_{\text{i}}\text{X}_{\text{i}}^2-\big(\sum\text{P}_{\text{i}}\text{X}_{\text{i}}\big)^2$
$=0\times\frac{256}{625}+1\times\frac{256}{625}+4\times\frac{96}{625}+9\times\frac{16}{625}+16\times\frac{1}{625}-\big(\frac{4}{5}\big)^2$
$=\frac{1}{625}(256+384+144+16)-\frac{16}{25}$
$=\frac{800}{625}-\frac{16}{25}$
$=\frac{400}{625}$
$=\frac{16}{25}$
Thus, the mean and vairance of the distribution are $\frac{4}{5}$ and $\frac{16}{25},$ respectively.
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Question 1694 Marks
There are three urns A, B, and C. Urn A contains 4 red balls and 3 black balls. urn B contains 5 red balls and 4 black balls. Urn C contains 4 red and 4 black balls. One ball is drawn from each of these urns. What is the probability that 3 balls drawn consists of 2 red balls and a black ball?
Answer
Urn A contains 4 red (R1) and 3 black (B1) balls
Urn B contains 5 red (R2) and 4 black (B2) balls
Urn C contains 4 red (R3) and 4 balck (B3) balls.
P (3 balls drawn consists or 2 red and a black ball)
$=\text{P}\big[(\text{R}_1\cap\text{R}_2\cap\text{R}_3)\cup(\text{R}_1\cap\text{B}_2\cap\text{R}_3)\cap(\text{B}_1\cap\text{R}_2\cap\text{R}_3)\big]$
$=\text{P}(\text{R}_1\cap\text{R}_2\cap\text{R}_3)+\text{P}(\text{R}_1\cap\text{B}_2\cap\text{R}_3)+\text{P}(\text{B}_1\cap\text{R}_2\cap\text{R}_3)$
$=\text{P}(\text{R}_1)\text{P}(\text{R}_2)\text{P}(\text{R}_3)+\text{P}(\text{R}_1)\text{P}(\text{B}_2)\text{P}(\text{R}_3)+\text{P}(\text{B}_1)\text{P}(\text{R}_2)\text{P}(\text{R}_3)$
$=\frac{4}{7}\times\frac{5}{9}\times\frac{4}{8}+\frac{4}{7}\times\frac{4}{9}\times\frac{4}{8}+\frac{3}{7}\times\frac{5}{9}\times\frac{4}{8}$
$=\frac{80+64+60}{504}$
$=\frac{204}{504}$
$=\frac{17}{42}$
Required probability $=\frac{17}{42}$
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Question 1704 Marks
Two bad eggs are accidently mixed up with ten good ones. Three eggs are drawn at random with replacement from this lot. Compute the mean for the number of bad eggs drawn.
Answer
Let X denote the number of bad eggs in a sample of 3 eggs drawn from a lot containing 2 bad eggs and 10 good eggs. Then, X can take the values 0, 1 and 2.

P(X = 0)

= P(no bad egg)

$=\frac{\text{}^{10}\text{C}_3}{\text{}^{12}\text{C}_3}$

$=\frac{120}{220}$

$=\frac{6}{11}$

P(X = 1)

= P(1 bad egg)

$=\frac{\text{}^{2}\text{C}_1\times\text{}^{10}\text{C}_2}{\text{}^{12}\text{C}_3}$

$=\frac{90}{220}$

$=\frac{9}{22}$

P(X = 2)

= P(2 bad eggs)

$=\frac{\text{}^{2}\text{C}_2\times\text{}^{10}\text{C}_1}{\text{}^{12}\text{C}_3}$

$=\frac{10}{220}$

$=\frac{1}{22}$

Thus, the probability distribution of X is given by

$\text{X}$

 $\text{P}(\text{X})$
$0$

 $\frac{6}{11}$
$1$

 $\frac{9}{22}$
$2$

 $\frac{1}{22}$
Computation of mean

$\text{x}_\text{i}$

 $\text{p}_\text{i}$

 $\text{x}_\text{i}\text{p}_\text{i}$
$0$

 $\frac{6}{11}$

 $0$
$1$

 $\frac{9}{22}$

 $\frac{9}{22}$
$2$

 $\frac{1}{22}$

 $\frac{1}{11}$

 

 

 $\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{1}{2}$
Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{1}{2}$

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Question 1714 Marks
In a group of 400 people, 160 are smokers and non-vegetarian, 100 are smokers and vegetarian and the remaining are non-smokers and vegetarian. The probabilities of getting a special chest disease are 35%, 20% and 10% respectively. A person is chosen from the group at random and is found to be suffering from the disease. What is the probability that the selected person is a smoker and non-vegetarian?
Answer
Let E1, E2, E3 be the events that the people are smokers and non-vegetarian, skokers and vegetarian, and non-smokers and vegetarian respectively.
$\text{P}(\text{E}_1)=\frac{2}{5},\text{P}(\text{E}_2)=\frac{1}{4},\text{P}(\text{E}_1)=\frac{7}{20}$
Let A denote the event that the person has the special chest disease. It is given that
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=0.35,\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=020,\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=0.10$
We have to find $\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)$
By Baye's theorem,
$\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$
$=\frac{\frac{2}{5}(0.35)}{\frac{2}{5}(0.35)+\frac{1}{4}(0.20)+\frac{7}{20}(0.10)}=\frac{\frac{7}{50}}{\Big(\frac{7}{50}\Big)+\Big(\frac{1}{20}\Big)+\Big(\frac{7}{200}\Big)}$
$=\frac{\frac{7}{50}}{\frac{9}{40}}=\frac{28}{45}$
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Question 1724 Marks
Find the mean, variance and standard deviation of the number of tails in three tosses of a coin.
Answer
Let X denotes the number of tails in three tosses of a coin. Then, X can take the values 0, 1, 2 and 3.

P(X = 0) = P(HHH) $=\frac{1}{8},$ P(X = 1) = P(THH or HHT or HTH) $=\frac{3}{8}$

P(X = 2) = P(TTH or THT or HTT) $=\frac{3}{8},$ P(X = 3) = P(TTT) $=\frac{1}{8}$

Thus, the probability distribution of X is given by

$0$

 $\frac{1}{8}$
$1$

 $\frac{3}{8}$
$2$

 $\frac{3}{8}$
$3$

 $\frac{3}{8}$
Computation of mean and step deviation

$\text{x}_\text{i}$

 $\text{p}_\text{i}$

 $\text{p}_\text{i}\text{x}_\text{i}$

 $\text{p}_\text{i}\text{x}_\text{i}^2$
$0$

 $\frac{1}{8}$

 $0$

 $0$
$1$

 $\frac{3}{8}$

 $\frac{3}{8}$

 $\frac{32}{221}$
$2$

 $\frac{3}{8}$

 $\frac{6}{8}$

 $\frac{4}{221}$
$3$

 $\frac{1}{8}$

 $\frac{3}{8}$  

 

 

 $\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{3}{2}$

 $\sum\text{p}_\text{i}\text{x}_\text{i}^2=3$
Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{3}{2}$

Mean $=\frac{34}{221}$

Variance $=\sum\text{p}_\text{i}\text{x}_\text{i}^2-(\text{Mean})^2$

$3-\Big(\frac{3}{2}\Big)^2$

$=\frac{3}{4}$

Step Deviation $=\sqrt{\text{Variance}}$

$=\sqrt{\frac{3}{4}}$

$=0.87$

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Question 1734 Marks
For the following probability distribution determine standard deviation of the
random variable X.
X
2
3
4
P(X)
0.2
0.5
0.3
Answer
We have
X
2
3
4
P(X)
0.2
0.5
0.3
XP(X)
0.4
1.5
1.2
X2P(X)
0.8
4.5
4.8
We know that, standard deviation of $\text{X}=\sqrt{\text{Var}(\text{X})}$
Where, $\text{var}(\text{X})=\text{E}(\text{X}^2)-\big[\text{E}(\text{X})\big]^2$
$=\sum_\limits{\text{i}=1}^\text{n}\text{x}^2_1\text{P}(\text{X}_1)-\Bigg[\sum_\limits{\text{i}=1}^\text{n}\text{x}_\text{i}\text{P}_{\text{i}}^2\Bigg]$
$\therefore\text{Var}(\text{X})=[0.8+4.5+4.8]-[0.4+1.5+1.2]^2$
$=10.1-(3.1)^2=10.1-9.61$
$=0.49$
$\therefore$ Standard deviation of $\text{X}=\sqrt{\text{Var}\text{X}}$
$=\sqrt{0.49}=0.7$
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Question 1744 Marks
Three cards are cdrawn successively with replacement from a well shffled deck of 52 cards. A random variable X denotes the number of hearts in the three cards drawn. Determine the probability distribution of X.
Answer
Three cards are thrown with replacement. Let X denote the numbers of hearts if three cards are drawn.
So, X has values 0, 1, 2, 3
$\text{P}(\text{X}=0)=\text{P}\big(\overline{\text{H}}_1\big)\times\text{P}\big(\overline{\text{H}}_2\big)\times\text{P}\big(\overline{\text{H}}_3\big)$
$=\frac{39}{52}\times\frac{39}{52}\times\frac{39}{52}$
$=\frac{27}{26}$
$\text{P}(\text{X}=1)=\text{P}(\text{H}_1)\text{P}\big(\overline{\text{H}}_2\big)\text{P}\big(\overline{\text{H}}_3\big)+\text{P}\big(\overline{\text{H}}_1\big)\text{P}(\text{H}_2)\text{P}\big(\overline{\text{H}}_3\big)+\text{P}\big(\overline{\text{H}}_1\big)\text{P}\big(\overline{\text{H}}_2\big)\text{P}(\text{H}_3)$
$=\frac{13}{52}\times\frac{39}{52}\times\frac{39}{52}+\frac{39}{52}\times\frac{13}{52}\times\frac{39}{52}+\frac{39}{52}\times\frac{13}{52}$
$=\frac{27}{26}$
$\text{P}(\text{X}=2)=\text{P}(\text{H}_1)\text{P}(\text{H}_2)\text{P}\big(\overline{\text{H}}_3\big)+\text{P}(\text{H}_1)\text{P}\big(\overline{\text{H}}_2\big)\text{P}(\text{H}_3)+\text{P}\big(\overline{\text{H}}_1\big)\text{P}(\text{H}_2)\text{P}(\text{H}_3)$
$=\frac{13}{52}\times\frac{13}{52}\times\frac{39}{52}+\frac{13}{52}\times\frac{39}{52}\times\frac{13}{52}+\frac{13}{52}\times\frac{13}{52}\times\frac{13}{52}$
$=\frac{9}{64}$
$\text{P}(\text{X}=3)=\text{P}(\text{H}_1)\text{P}(\text{H}_2)\text{P}(\text{H}_3)$
$=\frac{13}{52}\times\frac{13}{52}\times\frac{13}{52}$
$=\frac{1}{64}$
So,
Required probability distribution is
$\text{X}:$
$0$
$1$
$2$
$3$
$\text{P}(\text{X}):$
$\frac{27}{64}$
$\frac{27}{64}$
$\frac{9}{64}$
$\frac{1}{64}$
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Question 1754 Marks
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that,
  1. Both balls are red,
  2. First ball is black and second is red,
  3. One of them is black and other is red.
Answer
The Box contains 10 black balls and 8 red balls.

Then $\text{P(Black ball)}=\frac{10}{18}$

$\text{P(red ball)}=\frac{8}{18}$

  1. P(Both ballls are red) $=\frac{8}{18}\times\frac{8}{18}=\frac{16}{81}$

  2. P (First ball is black and second is red) $=\frac{10}{18}\times\frac{8}{18}=\frac{20}{81}$

  3. P (One of them is black and other is red)

$=\frac{10}{18}\times\frac{8}{18}+\frac{8}{18}\times\frac{10}{18}$

$=2\Big(\frac{20}{81}\Big)$

$=\frac{40}{81}$

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Question 1764 Marks
Three persons A, B and C apply for a job of Manager in a Private Company. Chances of their selection (A, B and C) are in the ratio 1 : 2 : 4. The probabilities that A, B and C can introduce changes to improve profits of the company are 0.8, 0.5 and 0.3, respectively. If the change does not take place, find the probability that it is due to the appointment of C.
Answer
Let E1, E2 and E3 be the events denoting the selecting of A, B and C as managers, respectively.
P(E1) = Ptobability of selection of A $=\frac{1}{7}$
P(E2) = Probability of selection of B $=\frac{2}{7}$
P(E3) = Probability of selection of C $=\frac{4}{7}$
Let be the event denoting the change not taking place.
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=$ Probability that A does not introduce change = 0.2
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=$ Probability that B does not introduce change = 0.5
$\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=$ Probability that C does not introduce change = 0.7
$\therefore$ Required probability $=\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)$
By Baye's theorem, we have
$\text{P}\Big(\frac{\text{E}_3}{\text{A}}\Big)=\frac{\text{P}(\text{E}_3)\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$
$=\frac{\frac{4}{7}\times0.7}{\frac{1}{7}\times0.2+\frac{2}{7}\times0.5+\frac{4}{7}\times0.7}$
$=\frac{2.8}{0.2+1+2.8}$
$=\frac{2.8}{4}=0.7$
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Question 1774 Marks
In a factory, machine A produces 30% of the total output, machine B produces 25% and the machine C produces the remaining output. If defective items produced by machines A, B and C are 1%, 1.2%, 2% respectively. Three machines working together produce 10000 items in a day. An item is drawn at random from a day's output and found to be defective. Find the probability that it was produced by machine B?
Answer
Consider events E1, E2, E3 and Aas:
E1 = Selecting product from machine A
E2 = Selecting product from machine B
E3 = Selecting product from machine C
A = Selecting a standard quality product
$\text{P}(\text{E}_1)=\frac{30}{100}$
$\text{P}(\text{E}_2)=\frac{25}{100}$
$\text{P}(\text{E}_3)=\frac{45}{100}$
P(A|E1) = P(Selecting defective product from machine A)
$=\frac{1}{100}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P}$ (Selecting defective prodcut from machine B)
$=\frac{1.2}{100}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=\text{P}$ (Selecting defective product from machine C)
$=\frac{2}{100}$
To find, P(Selecting defective product is produced by machine B)
By baye's theorem,
$\text{P}\Big(\frac{\text{E}_2}{\text{A}}\Big)=\frac{\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{ P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$
$=\frac{\frac{25}{100}\times\frac{12}{1000}}{\frac{30}{100}\times\frac{1}{100}+\frac{25}{100}\times\frac{12}{1000}+\frac{45}{100}\times\frac{2}{100}}$
$=\frac{300}{300+300+900}$
$=\frac{300}{1500}$
$=\frac{1}{5}$
Required probability $=\frac{1}{5}$
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Question 1784 Marks
Suppose 5 men out of 100 and 25 women out of 1000 are good orators. An orator is chosen at random. Find the probability that a male person is selected. Assume that there are equal number of men and women.
Answer
Given,
5 man out of 100 and 25 women out of 1000 are good orators.
Consider E1, E2 and A events as:
E1 = Selected persom is male
E2 = Selected person is famale
E3 = Selected person is an orator
$\text{P}(\text{E}_1)=\frac{1}{2}$
$\text{P}(\text{E}_2)=\frac{1}{2}$
[Since number of ,ales and females are equal]
P(A|E1) = P(Selecting a male orator)
$=\frac{5}{100}$
$=\frac{1}{20}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P}$ (Selecting a female orator)
$=\frac{25}{1000}$
$=\frac{1}{40}$
To find, P(Prator selected is a male) $=\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)$
By baye's theorem,
$\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{1}{2}\times\frac{1}{20}}{\frac{1}{2}\times\frac{1}{20}+\frac{1}{2}\times\frac{1}{40}}$
$=\frac{\frac{1}{40}}{\frac{1}{40}+\frac{1}{80}}$
$=\frac{1}{40}\times\frac{80}{3}$
$=\frac{2}{3}$
Required probability $=\frac{2}{3}.$
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Question 1794 Marks
A bag contains 4 white and 5 black balls and another bag contains 3 white and 4 black balls. A ball is taken out from the first bag and without seeing its colour is put in the second bag. A ball is taken out from the latter. Find the probability that the ball drawn is white.
Answer
There are two bags.
Bag (1) contain 4 white and 5 black balls.
Bag (2) contain 4 white and 4 black balls.
A ball is taken from bag (i) and without seeing its colout is put in second bag. Then a ball is drawn from bag 2 and is white in colour.
A (White ball from bag 1) $=\frac{4}{9}$
$\text{P}(\text{W}_1)=\frac{4}{9}$
P(Black ball from bag 1) $=\frac{5}{9}$
$\text{P}(\text{B}_1)=\frac{5}{9}$
P(white ball from bag 2 given B1 transfer)
$\text{P}\Big(\frac{\text{W}_2}{\text{B}_1}\Big)=\frac{3}{8}$
P(White from bag 2 given W1 transfer)
$\text{P}\Big(\frac{\text{W}_2}{\text{W}_2}\Big)=\frac{4}{8}$
$=\frac{1}{2}$
P(white from bag 2)
$=\text{P}(\text{B}_1)\text{P}\Big(\frac{\text{W}_2}{\text{B}_1}\Big)+\text{P}(\text{W}_1)\text{P}\Big(\frac{\text{W}_2}{\text{W}_1}\Big)$
$=\frac{5}{8}\times\frac{3}{8}+\frac{4}{9}\times\frac{1}{2}$
$=\frac{15}{72}+\frac{4}{18}$
$=\frac{31}{72}$
Required probability $=\frac{31}{72}$
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Question 1804 Marks
A and B throw a pair of dice alternately. A wins the game if he gets a total of 6 and B wins if she gets a total of 7. It A starts the game, find the probability of winning the game by A in third throw of the pair of dice.
Answer
A and B throw a pair of dice alternately.
A wins if he gets a total of 6
A = {(2, 4), (1, 5), (5, 1), (4, 2), (3, 3)}
B wins if she gets a total of 7
B = {(2, 5), (1, 6) (6, 1), (5, 2), (3, 4), (4,3)} 
Let P(A) is the probability, if A wins in a throw
$\Rightarrow\text{P}(\text{A})=\frac{5}{36}$
And P(B) is the probability, if B wins in a throw
$\Rightarrow\text{P}(\text{B})=\frac{1}{6}$
$\therefore$ Probability of winning the game by A in third throw
$=\text{P}(\bar{\text{A}})\cdot\text{P}(\bar{\text{B}})\cdot\text{P}(\text{A})$
$=\frac{31}{36}\cdot\frac{5}{6}\cdot\frac{5}{36}$
$=\frac{775}{216\cdot36}$
$=\frac{775}{7776}$
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Question 1814 Marks
The probability that a certain person will buy a shirt is 0.2, the probability that he will buy a trouser is 0.3, and the probability that he will buy a shirt given that he buys a trouser is 0.4. Find the probability that he will buy both a shirt and a trouser. Find also the probability that he will buy a trouser given that he buys a shirt.
Answer
Given,
Probability that a person buys a shirt (S) = P(S) = 0.2
Probability that he buys a trouser (T) = P(T) = 0.3
$\text{P}\Big(\frac{\text{S}}{\text{T}}\Big)=0.4$
We know that,
$\text{P}\Big(\frac{\text{S}}{\text{T}}\Big)=\frac{\text{P}(\text{S}\cap\text{T})}{\text{P}(\text{T})}$
$0.4=\frac{\text{P}(\text{S}\cap\text{T})}{0.3}$
$\text{P}(\text{S}\cap\text{T})=0.4\times0.3$
$\text{P}(\text{S}\cap\text{T})=0.12$
Probability that he buys a shirt and a trouser both = 0.12
$\text{P}\Big(\frac{\text{T}}{\text{S}}\Big)=\frac{\text{P}(\text{S}\cap\text{T})}{\text{P}(\text{S})}$
$=\frac{0.12}{0.2}$
$\text{P}\Big(\frac{\text{T}}{\text{S}}\Big)=\frac{12}{20}$
$=\frac{3}{5}$
$=0.6$
Probability that he buys a trouser given that he buys a shirt = 0.6
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Question 1824 Marks
A random variable X takes the values 0, 1, 2 and 3 such that:
P(X = 0) = P(X > 0) = P(X < 0); P(X = -3) = P(X = -2) = P(X = -1); P(X = 1) = P(X = 2) = P(X = 3).
Obtain the probability distribution of X.
Answer
Let P(X = 0) = k. Then,
P(X = 0) = P(X > 0) = P(X < 0)
⇒ P(X > 0) = k
P(X < 0) = k
$\therefore$ P(X = 0) + P(X > 0) + P(X < 0) = 1
⇒ k + k + k = 1
$\Rightarrow\text{k}=\frac{1}{3}$
Now,
P(X < 0) = k
⇒ P(X = -1) + P(X = -2) + P(X = -3) = k
⇒ 3P(X = -1) = k $[\because$ P(X = -1) = P(X = -2) = P(X = -3)$]$
$\Rightarrow\text{P}(\text{X}-1)=\frac{\text{k}}{3}$
$\Rightarrow\text{P}(\text{X}=-1)=\frac{1}{3}\times\frac{1}{3}=\frac{1}{9}$
$\therefore\ \text{P}(\text{X}=-1)=\text{P}(\text{X}=-2)=\text{P}(\text{X}=-3)=\frac{1}{9}$
Similarly,
P(X > 0) = k
$\Rightarrow\text{P}(\text{X}=1)=\text{P}(\text{X}=2)=\text{P}(\text{X}=3)=\frac{1}{9}$
Thus, the probability distribution is given by
$\text{X}_\text{i}$
 $\text{P}_\text{i}$
$-3$
 $\frac{1}{9}$
$-2$
 $\frac{1}{9}$
$-1$
 $\frac{1}{9}$
$1$
 $\frac{1}{9}$
$2$
 $\frac{1}{9}$
$3$
 $\frac{1}{9}$
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Question 1834 Marks
Suppose we have four boxes A, B, C, D containing coloured marbles as given below:

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from,

  1. Box A?
  2. Box B?
  3. Box C?
Answer
$\text{P}\Big(\frac{\text{A}}{\text{Red}}\Big),\text{P}\Big(\frac{\text{B}}{\text{Red}}\Big),\text{P}\Big(\frac{\text{C}}{\text{Red}}\Big)$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{Red}}\Big)=\frac{\text{P}\Big(\frac{\text{A}}{\text{Red}}\Big)\text{P(A)}}{\text{P}\Big(\frac{\text{A}}{\text{Red}}\Big)\text{P(A)}+\text{P}\Big(\frac{\text{B}}{\text{Red}}\Big)\text{P(B)}+\text{P}\Big(\frac{\text{C}}{\text{Red}}\Big)\text{P(C)}+\text{P}\Big(\frac{\text{D}}{\text{Red}}\Big)\text{P(D)}}$
$=\frac{\frac{1}{10}\times\frac{1}{4}}{\frac{1}{10}\times\frac{1}{4}+\frac{6}{10}\times\frac{1}{4}+\frac{8}{10}\times\frac{1}{4}+0}$
$=\frac{1}{1+6+8}=\frac{1}{15}$
Similarly,
$\text{P}\Big(\frac{\text{B}}{\text{Red}}\Big)=\frac{6}{15}$
$\text{P}\Big(\frac{\text{C}}{\text{Red}}\Big)=\frac{8}{15}$
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Question 1844 Marks
Find the probability of 4 turning up at least once in two tosses of a fair die.
Answer
Let p denote the 4 turning up in a toss of a fair die, So
$\text{p}=\frac{1}{6}$
$\text{q}=1-\frac{1}{6}$
$\text{q}=\frac{5}{6}$ [Since p + q = 1]
Let X denote the variable showing the number of turning 4 up in 2 tosses of die.
Probability of getting 4, r times in n tosses of a die is given by
$\text{p}(\text{X = r})=\text{ }^{\text{n}}\text{c}_\text{r}\text{p}^\text{r}\text{q}^{\text{n}-\text{r}}$
$=\text{ }^2\text{c}_\text{r}\big(\frac{1}{6}\big)^2\big(\frac{5}{6}\big)^{2-\text{r}}\dots(1)$
Probability of getting 4 at least once in tow tosses of a fair die
$=\text{P}(\text{X}=1)+\text{P}(\text{X}=2)$
$=1-\text{P}(\text{X}=0)$
$=1-\Big[\text{ }^2\text{c}_0\big(\frac{1}{6}\big)^0\big(\frac{5}{6}\big)^{2-0}\Big]$ [Using (1)]
$=1-\Big[1.1.\big(\frac{5}{6}\big)^2\Big]$
$=1-\Big[\frac{25}{36}\Big]$
$=\frac{36-25}{36}$
$=\frac{11}{36}$
So,
Required probability $=\frac{11}{36}$
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Question 1854 Marks
The odds against a certain event are 5 to 2 and the odds in favour of another event, independent to the former are 6 to 5. Find the probability that,
  1. At least one of the events will occur,
  2. None of the events will occur.
Answer
Given,

The adds against a certain event (say, A) are 5 to 2

$\Rightarrow\ \text{P}(\overline{\text{A}})=\frac{5}{5+2}$

$\text{P}(\overline{\text{A}})=\frac{5}{7}$

The odds in favour of another event (say, B) are 6 to 5

$\Rightarrow\ \text{P(B)}=\frac{6}{5+6}$

$\text{P(B)}=\frac{6}{11}$

$\text{P}(\overline{\text{B}})=1-\frac{6}{11}$

$\text{P}(\overline{\text{B}})=\frac{5}{11}$

  1. P (At least one of the events will occur)

= 1 - P (None of events occur)

$=1-\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$

$=1-\text{P}(\overline{\text{A}})\text{ P}(\overline{\text{B}})$

[Since events are independent]

$=1-\frac{5}{7}\times\frac{5}{11}$

$=1-\frac{25}{77}$

$=\frac{52}{77}$

Required probability $=\frac{52}{77}$

  1. P (None of the events will occur)

$=\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$

$=\text{P}(\overline{\text{A}})\text{ P}(\overline{\text{B}})$

$=\frac{5}{7}\times\frac{5}{11}$

$=\frac{25}{77}$

Required probability $=\frac{25}{77}$

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Question 1864 Marks
Three cards are drawn at random (without replacement) from a well shuffled pack of 52 cards. Find the probability distribution of number of red cards. Hence, find the mean of the distribution.
Answer
Let 'X' be the random variable which can assume values from 0 to 3.
P(X = 0)
$=\frac{\text{}^{26}\text{C}_3}{\text{}^{52}\text{C}_3}=\frac{2600}{22100}=\frac{2}{17}$
P(X = 1)
$=\frac{\text{}^{26}\text{C}_1\times\text{}^{26}\text{C}_2}{\text{}^{52}\text{C}_3}=\frac{8450}{22100}=\frac{13}{17}$
P(X = 2)
$=\frac{\text{}^{26}\text{C}_2\times\text{}^{26}\text{C}_1}{\text{}^{52}\text{C}_3}=\frac{8450}{22100}=\frac{13}{17}$
P(X = 3)
$=\frac{\text{}^{26}\text{C}_3}{\text{}^{52}\text{C}_3}=\frac{2600}{22100}=\frac{2}{17}$
Probability distribution of X:
$\text{X}=\text{x}_\text{i}$
$0$
$1$
$2$
$3$
$\text{P}(\text{X} =\text{x}_\text{i})$
$\frac{2}{17}$
$\frac{13}{34}$ 
$\frac{13}{34}$
$\frac{2}{17}$
$=\sum\limits_{\text{i=0}}^3(\text{x}_\text{i}\times\text{p}_\text{i})$
$=\text{x}_0\text{p}_0+\text{x}_1\text{p}_1+\text{x}_2\text{p}_2+\text{x}_3\text{p}_3$
$=0\times\frac{2}{17}+1\times\frac{13}{34}+2\times\frac{13}{34}+3\times\frac{2}{17}$
$=\frac{13+26+12}{34}$
$=\frac{51}{34}$
$=\frac{3}{2}$
$=1.5$
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Question 1874 Marks
Ten coins are tossed. What is the probability of getting at least 8 heads?
Answer
Let X is the random variable for getting a head.
Now $\text{P}(\text{X}=\text{r})={^\text{n}\text{C}_\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}} $ 
$\therefore\text{P}(\text{X}=\text{r})=\text{P}(\text{r}-8)+\text{P}(\text{r}-9)+\text{P}(\text{r}-10)$
$={^{10}}\text{C}_8\Big(\frac{1}{2}\Big)^{8}\Big(\frac{1}{2}\Big)^{10-8}+{^{10}}\text{C}_9\Big(\frac{1}{2}\Big)^{9}\Big(\frac{1}{2}\Big)^{10-9}+{^{10}}\text{C}_{10}\Big(\frac{1}{2}\Big)^{10}\Big(\frac{1}{2}\Big)^{10-10}$
$=\frac{10!}{8!2!}\cdot\Big(\frac{1}{2}\Big)^{10}+10\cdot\Big(\frac{1}{2}\Big)^{10}+1\cdot\Big(\frac{1}{2}\Big)^{10}$
$=\Big(\frac{1}{2}\Big)^{10}\Big[\frac{10\times9}{2}+10+1\Big]$
$=56\cdot\Big(\frac{1}{2}\Big)^{10}=\frac{7}{128}$
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Question 1884 Marks
A die is thrown 5 times. Find the probability that an odd number will come up exactly three times.
Answer
Let geting an odd number be a success in trial.
We have,
p = probability of getting an odd number in a trial $=\frac{3}{6}=\frac{1}{2}$
Also, $\text{q}=1-\text{p}=1-\frac{1}{2}=\frac{1}{2}$
Let X denote the number of success in a sample of 5 trials. Then,
X follows binomial distribution with parameters n = 5 and $\text{p = q}=\frac{1}{2}$
$\therefore\text{P(X = r})=\text{ }^{5}\text{C}_{\text{r}}\text{p}^{\text{r}}\text{q}^{(5-\text{r})}=\text{ }^5\text{C}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{(5-\text{r})}=\text{ }^5\text{C}_{\text{r}}\big(\frac{1}{2}\big)^5,$ wher r = 0, 1, 2, 3, 4, 5
Now,
Required probability = P(X = 3)
$=\text{ }^5\text{C}_3\big(\frac{1}{2}\big)^5$
$=\frac{10}{32}$
$=\frac{5}{16}$
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Question 1894 Marks
If A and B are two events such that,
$\text{P(A)}=\frac{1}{3},\text{P(B)}=\frac{1}{4}$ and $\text{P}(\text{A}\cap\text{B})=\frac{5}{12},$ then find P(A|B) and P(B|A).
Answer
We have,
$\text{P(A)}=\frac{1}{3},\text{P(B)}=\frac{1}{4}$ and $\text{P}(\text{A}\cup\text{B})=\frac{5}{12}$
As, $\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}=\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\text{P(A)}+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{1}{3}+\frac{1}{4}-\frac{5}{12}$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{2}{12}$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{1}{6}$
Now,
$\text{P}(\text{A}|\text{B})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\frac{\Big(\frac{1}{6}\Big)}{\Big(\frac{1}{4}\Big)}=\frac{4}{6}=\frac{2}{3}$ and
$\text{P}(\text{B}|\text{A})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}=\frac{\Big(\frac{1}{6}\Big)}{\Big(\frac{1}{3}\Big)}=\frac{3}{6}=\frac{1}{2}$
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Question 1904 Marks
A purse contains 2 silver and 4 copper coins. A second purse contains 4 silver and 3 copper coins. If a coin is pulled at random from one of the two purses, what is the probability that it is a silver coin?
Answer
Purse (I) contains (2) silver and 4 copper coins
Purse (II) contains 4 silver and 3 copper coins
Let E1, E2 and A are defined as
E1 = Selecting purse I
E2 = Selectin purse II
A = Drawinf a silver coin
$\text{P}(\text{E}_1)=\frac{1}{2}$
$\text{P}(\text{E}_2)=\frac{1}{2}$
[Since, there are only 2 purses]
P(A|E1) = P(A|silver coin from purse I)
$=\frac{2}{6}$
$​​=\frac{1}{3}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P}$ (A|silver coin from purse II)
By the law of total probability,
$\text{P(A)}=\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)$
$=\frac{1}{2}\times\frac{1}{3}+\frac{1}{2}\times\frac{4}{7}$
$=\frac{1}{6}+\frac{4}{14}$
$=\frac{7+12}{42}$
$=\frac{19}{42}$
Required probability $=\frac{19}{42}$
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Question 1914 Marks
A and B take turns in throwing two dice, the first to throw 10 being awarded the prize, show that if A has the first throw, their chance of winning are in the ratio 12 : 11.
Answer
Let E be the events of throwing 10 on a pair of dice,
E = {(4, 6), (5, 5), (6, 4)}
$\text{P(E)}=\frac{3}{37}$
$\text{P(E)}=\frac{1}{12}$
$\text{P}(\overline{\text{E}})=\frac{11}{12}$
A wins the game in first or 3rd or 5th throw, .....
Probability that A wins in first throw $\text{P(E)}=\frac{1}{12}$
Probability that A wins in 3rd throw
$=\text{P}(\overline{\text{E}})\text{P}(\overline{\text{E}})\text{P(E)}$
$=\Big(\frac{11}{12}\Big)^2\Big(\frac{1}{12}\Big)$
Probability that A wins in 5th throw
$=\text{P}(\overline{\text{E}})\text{P}(\overline{\text{E}})\text{P}(\overline{\text{E}})\text{P}(\overline{\text{E}})\text{P(E)}$
$=\Big(\frac{11}{12}\Big)^4\Big(\frac{1}{12}\Big)$
Hence,
Probability of winning A
$=\frac{1}{12}+\Big(\frac{11}{12}\Big)^2\Big(\frac{1}{12}\Big)+\Big(\frac{11}{12}\Big)^4\Big(\frac{1}{12}\Big)$
$=\frac{1}{12}\Big[1+\Big(\frac{11}{12}\Big)^2+\Big(\frac{11}{12}\Big)^4+\ .....\Big]$
$=\frac{1}{12}\bigg[\frac{1}{1-\big(\frac{11}{12}\big)^2}\bigg]\Big[\text{Since S}_{\infty}=\frac{\text{a}}{1-\text{r}}\ \text{for G. P.}\Big]$
$=\frac{1}{12}\bigg[\frac{1}{1-\frac{121}{144}}\bigg]$
$=\frac{1}{12}\times\frac{144}{23}$
$=\frac{12}{23}$
Probability of winning B
= 1 - P(Winning A)
$=1-\frac{12}{23}$
$=\frac{11}{23}$
Chances of winning A and B are $\frac{12}{23}$ and $\frac{11}{23}$ respectively or in 12 : 11.
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Question 1924 Marks
The probability that a student selected at random from a class will pass in Mathematics is $\frac{4}{5}$, and the probability that he/ she passes in Mathematics and Computer Science is $\frac{1}{2}$. What is the probability that he/ she will pass in Computer Science if it is known that he/ she has passed in Mathematics?
Answer
Given,
Probability to pass mathen atics (M)
$\text{P(M)}=\frac{4}{5}$
Probability to pass in mathematics (M) and computer Science (C)
$\text{P}(\text{M}\cap\text{C})=\frac{1}{2}$
To find, $\text{P}\Big(\frac{\text{C}}{\text{M}}\Big)$
We know tht,
$\text{P}\Big(\frac{\text{C}}{\text{M}}\Big)=\frac{\text{P}(\text{M}\cap\text{C})}{\text{P(M)}}$
$=\frac{\frac{1}{2}}{\frac{4}{5}}$
$=\frac{1}{2}\times\frac{5}{4}$
$=\frac{8}{5}$
Required probability $=\frac{8}{5}$
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Question 1934 Marks
If A and B are two events such that $2\text{P(A)}=\text{P(B)}=\frac{5}{13}$ and $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{2}{5}$ find $\text{P}(\text{A}\cap\text{B}).$
Answer
Given,
$2\text{P(A)}=\text{P(B)}=\frac{5}{13}$
$2\text{P(A)}=\frac{5}{13}$
$\Rightarrow \text{P(A)}=\frac{5}{26}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\frac{2}{5}=\frac{\text{P}(\text{A}\cap\text{B})}{\frac{5}{13}}$
$\text{P}(\text{A}\cap\text{B})=\frac{2}{5}\times\frac{5}{13}$
$\text{P}(\text{A}\cap\text{B})=\frac{2}{13}$
We know that,
$\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A})+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=\frac{5}{26}+\frac{5}{13}-\frac{2}{13}$
$=\frac{5+10-4}{26}$
$=\frac{11}{26}$
$\text{P}(\text{A}\cap\text{B})=\frac{11}{26}$
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Question 1944 Marks
Determine the binomial distribution whose mean is 9 and variance $\frac{9}{4}.$
Answer
Let X denote the variance with parameters n and p

$\text{p + q}=1$

$\text{q}=1-\text{p}$

Given,

$\text{Mean = np} =9\dots(1)$

$\text{variance = npq}=\frac{9}{4}\dots(2)$

$\frac{\text{npq}}{\text{np}}=\frac{\frac{9}{4}}{9}$ [By diving (1) by (2)]

$\text{q}=\frac{1}{4}$

So, $\text{p}=1-\text{q}$

$=1-\frac{1}{4}$

$\text{p}=\frac{3}{4}$

Put p in equation (1),

$\text{n}\big(\frac{3}{4}\big)=9$

$\Rightarrow\text{n}=\frac{36}{3}$

So, $\text{n}=12$

The distribution is given by

$=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}(\text{q})^{\text{n}-\text{r}}$

$\text{P(X = r})=\text{ }^{12}\text{c}_{\text{r}}\big(\frac{3}{4}\big)^{\text{r}}\big(\frac{1}{4}\big)^{12-\text{r}}$

$\text{for r}=0,1,2,\dots12$

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Question 1954 Marks
If A and B are two events such that,
$\text{P(A)}=\frac{6}{11},\text{P(B)}=\frac{5}{11}$ and $\text{P}(\text{A}\cap\text{B})=\frac{7}{11},$ then find $\text{P}(\text{A}\cap\text{B}),$ P(A|B) and P(B|A).
Answer
We have,
$\text{P(A)}=\frac{6}{11},\text{P(B)}=\frac{5}{11}$ and $\text{P}(\text{A}\cup\text{B})=\frac{7}{11}$
As, $\text{P}(\text{A}\cap\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{6}{11}+\frac{5}{11}-\frac{7}{11}$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{6+5-7}{11}$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{4}{11}$
Now,
$\text{P}(\text{A}|\text{B})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{b})}=\frac{\Big(\frac{4}{11}\Big)}{\Big(\frac{5}{11}\Big)}=\frac{4}{5}$ and
$\text{P}(\text{B}|\text{A})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{A})}=\frac{\Big(\frac{4}{11}\Big)}{\Big(\frac{6}{11}\Big)}=\frac{4}{6}=\frac{2}{3}$
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Question 1964 Marks
A pair of fair dice is thrown. Let X be the random variable which denotes the minimum of the two numbers which appear. Find the probability distribution, mean and variance of X.
Answer
A pair of fair dice is thrown. And X denote minimum of the two number appeared.

So, X can values 2, 3, 4, 5, 6.

$\text{P}(\text{X}=1)=\frac{11}{36}$ [Possible pairs: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1)]

$\text{P}(\text{X}=2)=\frac{9}{36}$ [Possible pairs: (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (4, 2), (5, 2), (6, 2)]

$\text{P}(\text{X}=3)=\frac{7}{36}$ [Possible pairs: (3, 3), (3, 4), (3, 5), (3, 6), (4, 3), (5, 3), (6, 3)]

$\text{P}(\text{X}=4)=\frac{5}{36}$ [Possible pairs: (4, 4), (4, 5), (4, 6), (5, 4), (6, 4)]

$\text{P}(\text{X}=5)=\frac{3}{36}$ [Possible pairs: (4, 4), (5, 5), (5, 6)5, (6, 5)]

$\text{P}(\text{X}=6)=\frac{1}{36}$ [Possible pairs: (6, 6)]

$\text{x}_\text{i}$

 $\text{p}_\text{i}$

 $\text{x}_\text{i}\text{p}_\text{i}$

 $\text{x}_\text{i}^2\text{p}_\text{i}$
$1$

 $\frac{11}{36}$

 $\frac{11}{36}$

 $\frac{11}{36}$
$2$

 $\frac{9}{36}$

 $\frac{18}{36}$

 $\frac{36}{36}$
$3$

 $\frac{7}{36}$

 $\frac{21}{36}$

 $\frac{63}{36}$
$4$

 $\frac{5}{36}$

 $\frac{20}{36}$

 $\frac{80}{36}$
$5$

 $\frac{3}{36}$

 $\frac{15}{36}$

 $\frac{75}{36}$
$6$

 $\frac{1}{36}$

 $\frac{6}{36}$

 $\frac{36}{36}$

 

 

 $\sum\text{xp}=\frac{91}{36}$

 $\sum\text{x}^2\text{p}=\frac{301}{36}$
Mean $=\sum\text{xp}$

Mean $=\frac{91}{36}$

Variance $=\sum\text{x}^2\text{p}-(\text{mean})^2$

$=\frac{301}{36}-\Big(\frac{91}{36}\Big)^2$

$=\frac{10836-8281}{1296}$

$=\frac{2555}{1296}$

Variance = 1.97

Probability distribution is

$\text{x}:$
$1$
$2$
$3$
$4$
$5$
$6$
$\text{p}(\text{x}):$
$\frac{11}{36}$
$\frac{9}{36}$
$\frac{7}{36}$
$\frac{5}{36}$
$\frac{3}{36}$
$\frac{1}{36}$
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Question 1974 Marks
Find the mean and variance of the number of tails in three tosses of a coin.
Answer
We know that in a throw of a coin.

$\text{P}(\text{H})=\frac{1}{2},\text{P}(\text{T})=\frac{1}{2}$

Let X denote the number of heads in three tosses of a coin.

So, X = 0, 1, 2, 3

P(X = 0) = P(T)P(T)P(T)

$=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}$

$=\frac{1}{8}$

P(X = 1) = P(H)P(T)P(T) + P(T)P(H)P(T) + P(T)P(T)P(H)

$=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}$

$=\frac{3}{8}$

P(X = 2) = P(H)P(H)P(T) + P(H)P(T)P(H) + P(T)P(H)P(H)

$=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}$

$=\frac{3}{8}$

P(X = 3) = P(H)P(H)P(H)

$=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}$

$=\frac{1}{8}$

So,

$\text{x}_\text{i}$

 $\text{p}_\text{i}$

 $\text{x}_\text{i}\text{p}_\text{i}$

 $\text{x}_\text{i}^2\text{p}_\text{i}$
$0$

 $\frac{1}{8}$

 $0$

 $0$
$1$

 $\frac{3}{8}$

 $\frac{3}{8}$

 $\frac{3}{8}$
$2$

 $\frac{3}{8}$

 $\frac{6}{8}$

 $\frac{12}{8}$
$3$

 $\frac{1}{8}$

 $\frac{1}{8}$

 $\frac{9}{8}$

 

 

 $\sum\text{xp}=\frac{3}{2}$

 $\sum\text{x}^2\text{p}$
Mean $=\sum\text{xp}=\frac{3}{2}$

Variance $=\sum\text{x}^2\text{p}-(\text{Mean})^2$

$=3-\frac{9}{4}=\frac{3}{4}$

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Question 1984 Marks
An urn contains m white and n black balls. A ball is drawn at random and is put back into the urn along with k additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. Show that the probability of drawing a white ball now does not depend on k.
Answer
Let Event, E1 = First ball drawn of white colour
Event, E2 = First ball drawn of black colour
And Events, E = Second ball drawn of white colour
$\therefore\text{P}(\text{E}_1)=\frac{\text{m}}{\text{m}+\text{n}}$ and $\text{P}(\text{E}_2)=\frac{\text{n}}{\text{m}+\text{n}}$
Also, $\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)=\frac{\text{m}+\text{k}}{\text{m}+\text{n}+\text{k}}$ and $\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)=\frac{\text{m}}{\text{m}+\text{n}+\text{k}}$
$\therefore$ Using total probability theorem,
$\text{P}(\text{E})=\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)$
$=\frac{\text{m}}{\text{m}+\text{n}}\cdot\frac{\text{m}+\text{k}}{\text{m}+\text{n}+\text{k}}+\frac{\text{n}}{\text{m}+\text{n}}\cdot\frac{\text{m}}{\text{m}+\text{n}+\text{k}}$
$=\frac{\text{m}(\text{m}+\text{k})+\text{nm}}{(\text{m}+\text{n}+\text{k})(\text{m}+\text{n})}$
$=\frac{\text{m}(\text{m}+\text{k}+\text{n})}{(\text{m}+\text{n}+\text{k})(\text{m}+\text{n})}$
$=\frac{\text{m}}{\text{m}+\text{n}}$
Hence, the probability of drawing a white ball does not depend on k.
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Question 1994 Marks
Suppose 10,000 tickets are sold in a lottery each for Rs. 1. First prize is of Rs. 3000 and the second prize is of Rs. 2000. There are three third prizes of Rs. 500 each. If you buy one ticket, what is your expectation.
Answer
Let x is the random variable for the prize.
$\text{X}$
 $0$
 $500$
 $2000$
 $3000$
$\text{P}(\text{X})$
 $\frac{9995}{10000}$
 $\frac{3}{10000}$
 $\frac{1}{10000}$
 $\frac{1}{10000}$
Since, $\text{E}(\text{X})=\sum\text{X}\ \text{P}(\text{X})$
$\therefore\text{E}(\text{X})=0\times\frac{9995}{10000}+\frac{1500}{10000}+\frac{2000}{10000}+\frac{3000}{10000}$
$=\frac{1500+2000+3000}{10000}$
$=\frac{6500}{10000}=\frac{13}{20}$
$=\text{Rs.}0.65$
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Question 2004 Marks
In a dice game, a player pays a stake of Rs. 1 for each throw of a die. She receives Rs. 5 if the die shows a 3, Rs. 2 if the die shows a 1 or 6, and nothing otherwise. What is the player’s expected profit per throw over a long series of throws?
Answer
Let X is the random variable of profit per throw.
$\text{X}$
 $-1$
 $1$
 $4$
$\text{P}(\text{X})$
 $\frac{1}{2}$
 $\frac{1}{3}$
 $\frac{1}{6}$
Since, she loses Rs. 1 On getting any of 2, 4 or 5.
So, at X = -1, $\text{P}(\text{X})=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}$
Similarly, at X = 1, $\text{P}(\text{X})=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}$ [if die shows of either 1 or 6]
and at X = 4, $\text{P}(\text{X})=\frac{1}{6}$ [if die shows a 3]
$\therefore$ Player's expected profit $=\sum(\text{X})=\sum\text{X}\text{P}(\text{X})$
$=-1\times\frac{1}{2}+1\times\frac{1}{3}+4\times\frac{1}{6}$
$=\frac{-3+2+4}{6}=\frac{1}{2}=\text{Rs.}0.50$
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